(Non)existence of circular arcs through set of points I have two circular arcs $ABCF$ and $ADEF$ which have the same endpoints $A, F$ and 'contain' the points $B, C$ and $D, E$ in the specified order.
How can I prove that there cannot be a circular arc $BCDE$, i.e. one that starts in $B$, intersects $C$ before $D$ and ends in $E$?

If both arcs are minor circular arcs, I can show that $B$ and $E$ must lie on different sides of the line through $C$ and $D$, hence either $BCD$ would be clockwise and $CDE$ counterclockwise or the other way around. Either way they could not form a circular arc $BCDE$ together.


However the case where one (or both) are major circular arcs cause me trouble.
Edit 1: Please note that the circular arcs may not overlap.
Edit 2: To clarify – the 'drawing' whose (non)existence I want to prove contains circular arcs $AB, BC, CF, AD, DE, EF, BC, CD, DE$. None of these may overlap and they shall form larger circular arcs together: $ABCF$ is made of $AB, BC, CF$; $ADEF$ is made of $AD, DE, EF$; and $BCDE$ is made of $BC, CD, DE$.
 A: This answer calls figure 1 below "a quadrilateral $BCDE$" and figure 2 below "a quadrilateral $BCED$". The vertices of figure 1 are $B,C,D,E$ clockwise. On the other hand, the vertices of figure 2 are $B,C,E,D$ clockwise.
$\qquad\qquad\qquad$

To prove by contradiction that there is no cirlce on which $B,C,D,E$ exist in this order, we suppose that there is such a circle.
Let us see, under the supposition, what conditions the four point $B,C,D,E$ in the figure given in the question has to satisfy. 
Now, suppose that there is a circle on which $B,C,D,E$ exist in this order. (see the figure below. Note that the quadrilateral here is $BCDE$, not $BCED$.) 
$\qquad\qquad\qquad$
Then, from the inscribed angle theorem, we get
$$\angle{BCE}=\angle{BDE}\tag1$$
$$\angle{CBD}=\angle{CED}\tag2$$
$$\angle{CDB}=\angle{CEB}\tag3$$
$$\angle{DCE}=\angle{DBE}\tag4$$
These are the necessary conditions which the four points $B,C,D,E$ in the figure given in the question has to satisfy to have a circle on which $B,C,D,E$ exist in this order.
Now let us come back to the figure given in the question. 
We now know that the four points $B,C,E,D$ in the figure given in the question have to satisfy $(1)(2)(3)(4)$. 
So, from $(1)(2)$, we have that the quadrilateral $BCED$ (not $BCDE$) has to be a parallelogram.
And from $(3)(4)$, the four inner angels have to be the same and so we have that the quadrilateral $BCED$ (again, not $BCDE$) has to be a rectangle. (see the figure below. The figure below is the same as the figure given in the question except that we have that the quadrilateral $BCED$ has to be a rectangle.)
$\qquad\qquad$
Then, finally, we see that the order of the four points on the circumscribed circle of the rectangle $BCED$ has to be $B,C,E,D$. 
This contradicts the supposition that there is a circle on which $B,C,D,E$ exist in this order.
It follows from this that there is no circle on which $B,C,D,E$ exist in this order. $\blacksquare$
A: Suppose we have the unit circle centered at the origin. Then we can assign points $A$, $B$, $C$, $D$, $E$, and $F$ have coordinates
$$A = \left( \cos \theta_1, \sin \theta_1 \right), $$
$$B = \left( \cos \theta_2, \sin \theta_2 \right), $$
$$C = \left( \cos \theta_3, \sin \theta_3 \right), $$
$$D = \left( \cos \theta_4, \sin \theta_4 \right), $$
$$E = \left( \cos \theta_5, \sin \theta_5 \right), $$
$$F = \left( \cos \theta_6, \sin \theta_6 \right), $$
where $$ \theta_1 < \theta_4 < \theta_5 < \theta_6 < \theta_3 < \theta_2.$$
Does this sort of reasoning take you any closer?
A: There is only one circle circumscribed circle which have for center the intercection of the bissection of the triangle.
The intersection of the bissections of a triangle is inside the Area of a triangle so it can not belong to both triangles !
